package com.sxkiler.demo.hard;

import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
import java.util.*;
import com.sxkiler.demo.model.*;

/**
count-of-range-sum=区间和的个数
<p>给定一个整数数组&nbsp;<code>nums</code>，返回区间和在&nbsp;<code>[lower, upper]</code>&nbsp;之间的个数，包含&nbsp;<code>lower</code>&nbsp;和&nbsp;<code>upper</code>。<br>
区间和&nbsp;<code>S(i, j)</code>&nbsp;表示在&nbsp;<code>nums</code>&nbsp;中，位置从&nbsp;<code>i</code>&nbsp;到&nbsp;<code>j</code>&nbsp;的元素之和，包含&nbsp;<code>i</code>&nbsp;和&nbsp;<code>j</code>&nbsp;(<code>i</code> &le; <code>j</code>)。</p>

<p><strong>说明:</strong><br>
最直观的算法复杂度是&nbsp;<em>O</em>(<em>n</em><sup>2</sup>) ，请在此基础上优化你的算法。</p>

<p><strong>示例:</strong></p>

<pre><strong>输入: </strong><em>nums</em> = <code>[-2,5,-1]</code>, <em>lower</em> = <code>-2</code>, <em>upper</em> = <code>2</code>,
<strong>输出: </strong>3 
<strong>解释: </strong>3个区间分别是: <code>[0,0]</code>, <code>[2,2]</code>, <code>[0,2]，</code>它们表示的和分别为: <code>-2, -1, 2。</code>
</pre>

 */
public class countRangeSum {
    

    class Solution {
        public Integer countRangeSum(Integer[] param0,Integer param1,Integer param2) {
            return null;
        }
    }

    @Test
    public void test(){
        Solution solution = new Solution();
        /**
        [-2,5,-1]
-2
2
        */
        //int [] num1 = new int[]{1,3};
        //int [] num2 = new int[]{2};
        //Assertions.assertEquals(solution.{{questionName}}(num1,num2),2);
    }
}

